Define $f(x, y, z) = \sin(y) + xz$. Let $\vec{a} = (3, \pi, -5)$ and $\vec{v} = \left( -2, 0, 3 \right)$. Calculate $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h}$.
Our ultimate goal is to substitute $h = 0$ into the limit and get the answer. To do this, we first need to cancel out the $h$ in the denominator. Otherwise we're dividing by zero. How can we simplify the limit so that we get an $h$ on the top and an $h$ on the bottom, which we can then cancel? Let's plug in $\vec{a}$ and $\vec{v}$ to the limit. $ \lim_{h \to 0} \dfrac{f \left( (3, \pi, -5) + h \left( --2, 0, 3\right) \right) - f(3, \pi, -5)}{h}$ Now we can add together the vectors. $ \lim_{h \to 0} \dfrac{f \left( 3 - 2h, \pi, -5 + 3h \right) - f(3, \pi, -5)}{h}$ Let's evaluate $f$. $ \lim_{h \to 0} \dfrac{\sin(\pi) + (3 - 2h)(-5 + 3h) - (\sin(\pi) + 3 (-5))}{h}$ We can cancel out all the terms without an $h$. $ \lim_{h \to 0} \dfrac{-15 + 9h + 10h - 6h^2 + 15}{h}$ becomes $ \lim_{h \to 0} \dfrac{19h - 6h^2}{h}$ Now we can cancel $h$ and calculate the limit. $\begin{aligned} \lim_{h \to 0} \dfrac{19h - 6h^2}{h} &= \lim_{h \to 0} 19 - 6h \\ \\ &= 19 \end{aligned}$ In conclusion, $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h} = 19$.